#### Answer

The areas of the four resulting smaller triangles is one fourth of the area of the given rhombus.

#### Work Step by Step

Draw diagonals MP and QN. the intersection of diagonals is named as O.
Let us compare $\triangle$ OPN and $\triangle$ OQP
we know in rhombus diagonals are perpendicular bisectors to each other
So OP = OM and OQ = ON
$\angle$ PON = $\angle$ POQ = $\angle$ MON = $\angle$ MOQ = 90$^{\circ}$
In $\triangle$ OPN and $\triangle$ OQP
PN =QP(in rhombus opposite sides are equal)
OP = OP (common)
$\angle$ PON = $\angle$ POQ = 90$^{\circ}$
By SAS rule $\triangle$ OPN is congruent to $\triangle$ OQP
Therefore area of $\triangle$ OPN = area of $\triangle$ OQP
Since In rhombus sides are equal
Similarly
area of $\triangle$ OMN = area of $\triangle$ OMQ
area of $\triangle$ PON = area of $\triangle$ OMN
area of $\triangle$ OQP = area of $\triangle$ OQM
so
area of $\triangle$ OQP = area of $\triangle$ OPN = area of $\triangle$ OQM = area of $\triangle$ OQN
Therefore area of MNPQ = area of $\triangle$ OQP + area of $\triangle$ OPN + area of $\triangle$ OQM + area of $\triangle$ OQN.
Therefore area of MNPQ = 4 * area of $\triangle$ OQP
area of $\triangle$ OQP = $\frac{area of MNPQ}{4}$.
Therefore area of smaller triangle is one fourth of the area of rhombus MNPQ.